Difference between revisions of "Arcsin"

From specialfunctionswiki
Jump to: navigation, search
Line 40: Line 40:
 
<div class="toccolours mw-collapsible mw-collapsed">
 
<div class="toccolours mw-collapsible mw-collapsed">
 
<strong>Proposition:</strong>  
 
<strong>Proposition:</strong>  
$\mathrm{arcsin}(z)=\sum_{k=0}^{\infty} \dfrac{\left(\frac{1}{2} \right)_n}{(2n+1)n!}x^{2n+1}$
+
$\mathrm{arcsin}(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{\left(\frac{1}{2} \right)_n}{(2n+1)n!}x^{2n+1}$
 
<div class="mw-collapsible-content">
 
<div class="mw-collapsible-content">
 
<strong>Proof:</strong> █  
 
<strong>Proof:</strong> █  

Revision as of 05:57, 31 October 2014

The function $\mathrm{arcsin} \colon [-1,1] \rightarrow \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$ is the inverse function of the sine function.

Properties

Proposition: $\dfrac{d}{dz} \mathrm{arcsin(z)} = \dfrac{1}{\sqrt{1-z^2}}$

Proof: If $\theta=\mathrm{arcsin}(z)$ then $\sin(\theta)=z$. Now use implicit differentiation with respect to $z$ to get $$\cos(\theta)\theta'=1.$$ The following image shows that $\cos(\mathrm{arcsin}(z))=\sqrt{1-z^2}$:

Cos(arcsin(z)).png

Hence substituting back in $\theta=\mathrm{arccos}(z)$ yields the formula $$\dfrac{d}{dz} \mathrm{arcsin(z)} = \dfrac{1}{\cos(\mathrm{arcsin(z)})} = \dfrac{1}{\sqrt{1-z^2}}. █$$

Proposition: $\int \mathrm{arcsin}(z) dz = \sqrt{1-z^2}+z\mathrm{arcsin}(z)+C$

Proof:

Proposition: $\mathrm{arcsin}(z) = \mathrm{arccsc}\left( \dfrac{1}{z} \right)$

Proof:

Proposition: $\mathrm{arcsin}(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{\left(\frac{1}{2} \right)_n}{(2n+1)n!}x^{2n+1}$

Proof:


References